Answer: According to recentest research no well actually it'd float away
Step-by-step explanation:
No, a lone variable has an invisible one in front of it, So 4z (can’t put the top number sorry) So 4z - z would equal 3z
The cost can be optimized by using a Linear Programming given the linear constraint system
- To minimize the cost, the biologist should use <u>60 samples of Type I</u> bacteria and <u>0 samples of Type II</u> bacteria
Reason:
Let <em>X</em> represent Type 1 bacteria, and let <em>Y</em>, represent Type II bacteria, we have;
The constraints are;
4·X + 3·Y ≥ 240
20 ≤ X ≤ 60
Y ≤ 70
P = 5·X + 7·Y
Solving the inequality gives;
4·X + 3·Y ≥ 240
(Equation for the inequality graphs)
The boundary of the feasible region are;
(20, 70)
(20, 53.
)
(60, 0)
(60, 70)
The cost are ;
![\begin{array}{|c|c|c|}X&Y&P= 5\times X + 7 \times Y\\20&70&590\\20&53.\overline 3&473.\overline 3\\60&0&300\\60&70&790\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7Cc%7C%7DX%26Y%26P%3D%205%5Ctimes%20X%20%2B%207%20%5Ctimes%20Y%5C%5C20%2670%26590%5C%5C20%2653.%5Coverline%203%26473.%5Coverline%203%5C%5C60%260%26300%5C%5C60%2670%26790%5Cend%7Barray%7D%5Cright%5D)
- Therefore, the minimum cost of $300 is obtained by using <u>60 samples of Type I</u> and <u>0 samples of Type II</u>
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Answer:
<h2>37</h2>
Step-by-step explanation:

p = 10
r = -4
t = 5
Substitute values into the given equation

Simplify

Your answer would be D
The work: 0.77 times 6
and 2.95 times 2
those equal $4.62 and $5.9
add those two together, and you'll have $10.52